consider 2 inertial reference frames, A and B, where B (a train) is moving with constant velocity, 10kn/h, in the positive x direction with respect to A.(a person standing on the station platform). if another person is throwing a ball on the train (ie in frame B at a speed of 10m/s in the negative y-direction in the train (ie frame B ), what is the measured velocity of the ball and direction in frame A (ie with respect to the platform xy)?
The ball has a downward velocity component if 10 m/s in both reference frames. In addition to that, it has a velocity component + 10m/s in the station platform frame (A).
So, take the vector sum of 10 m/s horizontal and -10 km/h vertical velocity components
why is it -10 km/h for the vertical velocity components?because i thought the train is travelling horizontally at 10km/h so it should be a horizaontal component?
<<why is it -10 km/h for the vertical velocity components? >>
Because the ball is thrown down. The horizontal relative motion of the two reference frames does not affect vertical velocity componnts when the transformation is performed.
To determine the measured velocity of the ball and its direction in frame A, we first need to consider the relative velocities between the two frames and then apply the principle of vector addition.
In this scenario, Frame B (the moving train) has a velocity of 10 km/h in the positive x-direction with respect to Frame A (the station platform). Symbolically, we can represent this as Bv/A = +10 km/h i, where i is the unit vector in the x-direction.
Next, we need to determine the velocity of the ball in Frame B relative to Frame A. The ball's velocity in Frame B is given as 10 m/s in the negative y-direction. Symbolically, we can represent this as Bv/B = -10 m/s j, where j is the unit vector in the y-direction.
To find the velocity of the ball in Frame A, we need to add the velocities together. Using vector addition, we have:
Av/A = Bv/A + Bv/B
Let's break it down step by step:
1. Convert the velocity of the train from km/h to m/s:
Bv/A = (10 km/h) x (1000 m/km) / (3600 s/h) = 10000 m/3600 s i = (10000/3600) m/s i
2. Add the velocities:
Av/A = (10000/3600) m/s i - 10 m/s j
So, the measured velocity of the ball in Frame A is (10000/3600) m/s i - 10 m/s j. This means that the ball appears to move in the negative y-direction and slightly in the positive x-direction from the perspective of someone in Frame A (the station platform).